Ruby hypothetical - variable nested loops

Question

This is just a thought exercise and I'd be interested in any opinions. Although if it works I can think of a few ways I'd use it.

Traditionally, if you wanted to perform a function on the results of a nested loop formed from arrays or ranges etc, you would write something like this:

def foo(x, y)
  # Processing with x, y
end

iterable_one.each do |x|
  iterable_two.each do |y|
      my_func(x, y)
  end
end

However, what if I had to add another level of nesting. Yes, I could just add an additonal level of looping. At this point, let's make foo take a variable number of arguments.

def foo(*inputs)
  # Processing with variable inputs
end

iterable_one.each do |x|
  iterable_two.each do |y|
    iterable_three.each do |z|
      my_func(x, y, x)
    end
  end
end

Now, assume I need to add another level of nesting. At this point, it's getting pretty gnarly.

My question, therefore is this: Is it possible to write something like the below?

[iterable_one, iterable_two, iterable_three].nested_each(my_func)

or perhaps

[iterable_one, iterable_two, iterable_three].nested_each { |args| my_func(args) }

Perhaps passing the arguments as actual arguments isn't feasible, could you maybe pass an array to my_func, containing parameters from combinations of the enumerables?

I'd be curious to know if this is possible, it's probably not that likely a scenario but after it occurred to me I wanted to know.

steenslag · Accepted Answer · 2013-02-27 19:28:53Z

4

Array.product yields combinations of enums as if they were in nested loops. It takes multiple arguments. Demo:

a = [1,2,3]
b = %w(a b c)
c = [true, false]

all_enums = [a,b,c]
all_enums.shift.product(*all_enums) do |combi|
  p combi
end


#[1, "a", true]
#[1, "a", false]
#[1, "b", true]
#...

answered Feb 27, 2013 at 19:28

steenslag

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2 Comments

Pete Hamilton Over a year ago

Facepalm. Totally forgot about product.

Pete Hamilton Over a year ago

I even used it for something the other day. Oh well, at least that's my curiosity satisfied. Early night clearly needed...

Rahul Tapali · Accepted Answer · 2013-02-27 19:30:10Z

3

You can use product:

[1,4].product([5,6],[3,5])  #=> [[1, 5, 3], [1, 5, 5], [1, 6, 3], [1, 6, 5], [4, 5, 3], [4, 5, 5], [4, 6, 3], [4, 6, 5]]

answered Feb 27, 2013 at 19:30

Rahul Tapali

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